Unit 4: Turning Effects of Force — Long Questions

Like Parallel Forces If the parallel forces are acting in the same direction, then they are called like parallel forces.

Unlike Parallel Forces If the parallel forces are acting in the opposite direction, then they are called like parallel forces.

Example Consider three forces F1, F2, and F3 acting on a rigid body at different points, as shown in Fig. 4.1. Here, the forces F1 and F2 are like parallel forces because they act in the same direction. In contrast, F2 and F3 are unlike parallel forces because they act in opposite directions.

Resultant Force A resultant force is a single force that has the same effect as the combined effect of all the forces to be added.

Head to tail rule The head-to-tail rule is a method used to add two or more Vectors, like forces. To apply this rule, place the tail of the second vector at the head of the first vector. If there are more than two vectors, continue placing the tail of each next vector at the head of the previous vector. After arranging all the vectors in this way, the resultant vector is drawn from the tail of the first vector to the head of the last vector. This resultant represents the total effect of all the vectors that are added.

Example 4.1

Let us add three force vectors F1, F2 and F3 having magnitudes of 200 N, 300 N and 250 N acting at angles of 30°, 45°, 60° with x - axis.(Shown in Fig.) By selecting a suitable scale 100 N = 1 cm, we can draw the force vectors as shown in Fig.(a). To add these vectors, we apply head-to-tail rule as shown in Fig.(b).

Measured length of resultant force is 7.1 cm. according to selected scale, magnitude of the resultant force F is 710 N and direction is at an angle 43° with x.- axis. (Shown in Fig.)

Rigid body If the distance between two points of the body remains the same under the action of a force, it is called a rigid body. A rigid body is the one that has no deformation by applying force.

Axis of rotation During rotation, all the particles of the rigid body rotate along fixed circles. The straight line joining the centers of these circles is called the axis of rotation in this case, it is OZ.

Line of action of force:
The line along which the force acts is called the line of action of the force.

Moment Arm The perpendicular distance from the axis of rotation to the line of action of the force is known as the moment arm. A larger moment arm results in a greater turning effect.

Moment of Force (Torque) The turning effect of a force is measured by a quantity known as moment of force or torque. Moment of a force or torque is defined as the product of the force and the moment arm.

Calculation of Torque The magnitude of torque is given by the formula:

τ = F × l

Where τ (tau) is the torque, F is the force, and l is the moment arm. In the case where the line of action of a force F is perpendicular to r, the moment arm l is equal to r. (Shown in Fig.)

Zero Torque Condition The torque of a force is zero when the line of action of a force passes through the axis of rotation, because its moment arm becomes zero.
τ = F × r = F × 0 = 0 as r = 0

Direction of Torque The torque is positive if the force tends to produce an anticlockwise rotation about the axis, and it is taken as negative if the force tends to produce a clockwise rotation.

Unit of Torque The SI unit of torque is newton meter (N·m).

Handling Non-Perpendicular Forces

In many cases, the line joining the axis of rotation and point P where the force F acts, is not perpendicular to the force F. Therefore, OP will not be the moment arm for F. In such cases, we have to find a component of force F that is perpendicular to OP = l (Fig.4.2), or we can find r, the component of l that is perpendicular to the line of action of force F (Fig.4.3). To resolve this, we need to know how to find the rectangular components of a force or any vector, a process known as the resolution of forces.

Couple When two equal and opposite parallel forces act at two different points of the same body, they form a couple.

Explanation The two forces are equal in magnitude but opposite in direction. Because they are applied at different points, they produce a turning effect (torque) on the object.

Examples from Daily Life

• Opening or closing a water tap
• Turning a key in a lock
• Opening the lid of a jar
• Turning the steering wheel of a motor car

Resolution of Vectors/ forces By the head-to-tail rule, two or more vectors can be added to give a resultant vector. The reverse process is also possible: a given vector can be divided into two or more parts called components. If these components are added, their resultant equals the original vector. Dividing a force into its components is known as the resolution of a force.

Perpendicular/ Rectangular Components Usually, a force is resolved into two perpendicular components, known as rectangular components. These components represent the force's effective values in the horizontal (x-axis) and vertical (y-axis) directions.

Process of Resolution Let us resolve a force F into its perpendicular components. A force F acting on a body at an angle θ with the x-axis is shown in Fig. 4.4a. Imagine a beam of light is placed above the vector F. As the light falls perpendicularly to the x-axis, it will cast a shadow of vector F onto the x-axis. We call this shadow the x-component of vector F.

Component A component of a vector is its effective value in a given direction.

Determining Components The x and y components can be practicaly drawn by dropping perpendiculars from the tip of vector F onto the x and y axes, respectively. The x-component of force F is denoted as Fx and the y-component as Fy. The magnitudes of the perpendicular components can be found from the right-angled triangle OAC in Fig. 4.4b:

X-Component (Fx)

In the given △OAC Base / Hyponenuse = cosθ
OA / OC = cosθ
Fx / F = cosθ
or Fx = Fcosθ ............ (4.2)

Fx is called horizontal component.

Y-Component (Fy)

In the given △OAC Perpendicular / Hyponenuse = sinθ
AC / OC = sinθ
Fy / F = sinθ
or Fy=Fsinθ

Fy is called vertical component.

The magnitude and direction of a force can be found if its perpendicular components are known. Consider a right-angled triangle OAC (Fig. 4.4b), where Fx and Fy are the perpendicular components of the force F.

Magnitude of the Force Applying the Pythagorean theorem to the right-angled triangle:
(hyp)² = (Base)² + (Prep)²
(OC)²=(OA)²+(AC)²
In terms of the components of force:
F²=Fx²+Fy²

Therefore, the magnitude F of the force is: F= √(Fx² + Fy²)

Direction of the Force The direction θ of the force F is given by the relationship between its components Fx and Fy:

tanθ = Perpendicular / Base
tanθ = Fy / Fx

Hence, the angle θ is θ = tan⁻¹(Fy / Fx)

Principle of Moments The principle of moments states that when a body is in balanced position, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.

Example Balance a metre rule on a wedge at its center of gravity (CG) such that the meter rule stays horizontal. Then suspend two weights, w1 and w2, on one side of the metre rule at distances l1 and l2 from the center of gravity. Place a third weight, w3, on the other side at a distance l3 from the center. The weights w1 and w2 will tend to rotate the metre rule anticlockwise about the center of gravity (CG).The weight w3 will tend to rotate it clockwise.

Moment Calculation The values of the moments of the weights are:
Moment of w1= w1× l1
Moment of w2= w2× l2
Moment of w3= w3× l3
When the metre rule is balanced:
Total anticlockwise moments = Total clockwise moments
w1 × l1+ w2× l2= w3× l3

Centre of gravity "Centre of gravity is that point where total weight of the body appears to be acting" If a body is supported at its center of gravity, it remains balanced without rotating.

An object is composed of a large number of small particles, and each of these particles experiences a gravitational force directed towards the center of the Earth. Since the object is small compared to the Earth, the value of gravitational acceleration 'g' can be considered uniform for all the particles. As a result, each particle experiences the same force mg, where (m) is the mass of the particle.

Since all these forces are parallel and act in the same direction, their resultant force will be the sum of all these individual forces, i.e.,
Resultant force = Σmg = ΣF = W
This sum of gravitational forces is equal to the total weight of the object w=Σmg, where M=Σm is the total mass of the object.

Center of Gravity of an Irregular-Shaped Plane Lamina:
For an irregular-shaped plane lamina, the center of gravity can be found by suspending it freely from different points. Every time the lamina is suspended, its center of gravity lies along the vertical line drawn from the suspension point, which is located using a plumb line. The exact position of the center of gravity is at the point where the two vertical lines, drawn from different suspension points, intersect. The center of gravity can exist either inside or outside the body. For instance, the center of gravity of a cup lies outside the object.

Center of Mass The center of mass of a body is that point where the whole mass of the body is assumed to be concentrated.

Relation with Center of Gravity:
On the surface of the Earth, where gravitational acceleration (g) is nearly uniform, the center of mass coincides with the center of gravity. This means the point at which all the mass is assumed to be concentrated (center of mass) is also the point where the total gravitational force acts.

Equilibrium

We know that if a number of forces act on a body such that their resultant is zero, the body remains at rest or continues to move with uniform velocity if already in motion. This state of the body is known as equilibrium. It can be stated as:
"A body is said to be in equilibrium if it has no acceleration."

Types of Equilibrium

Equilibrium is of two types:

1. Static Equilibrium:

A body at rest is in static equilibrium. In this state, the sum of all the forces acting on the body is zero.

Example

A book lying on a table is an example of static equilibrium. Only two forces act on it:
Weight (W=mg) acting downward.
Normal force (Fn) acting upward.
Since the book is at rest and has zero acceleration:
Fn-W=0 or Fn=W

Other examples include an electric bulb hanging from the ceiling, a man holding a box, and a beam held horizontally against a wall with the help of a rope.

2. Dynamic Equilibrium:

A body moving with uniform velocity is in dynamic equilibrium. In this case, the forces acting on the body balance each other, and there is no acceleration.

Example

A paratrooper descending with a uniform velocity after the parachute opens. In this state, the downward force of gravity acting on the paratrooper is balanced by the upward air resistance:
Fgravity = Fair resistance

Conditions of Equilibrium

For a body to be in equilibrium, two main conditions must be satisfied: the first condition relates to translational motion, and the second condition relates to rotational motion.

First Condition of Equilibrium

According to Newton's second law of motion F =ma
If a body is in translational equilibrium, then a=0, therefore, the net force F acting on the body must also be zero.

Mathematical Expression

ΣF=0.............. (1)

Statement

"A body is said to be in translational equilibrium only if the vector sum of all the external forces acting on it is equal to zero."

Resolving Forces into Components

In the case of coplanar forces (F1, F2, F3 , etc.), this condition can be broken into rectangular components:F1, F2, F3

Along the x-axis

F1x+F2x+F3x+···=0 Or ΣFx=0

Along the y-axis

F1y+F2y+F3y+···=0 Or ΣFy=0

"The sum of all the components of forces along the x-axis should be zero, and the sum of all the components of forces along the y-axis should also be zero." In other words we can say that sum of all the forces acting on the body is equal to zero. i.e. Σ F = 0

Second Condition of Equilibrium

This condition applies to rotational equilibrium which means that the body should not rotate under the action of the forces.

To solve problems by applying conditions of equilibrium, the following steps will help:

1. Select the Objects: First of all, select the objects to which ΣFx=0 and ΣFy=0 is to be applied. Each object should 'be treated separately.

2. Draw a Diagram: Draw a diagram to show the objects and forces acting on them. Only the forces acting on the objects should be included. The forces which the objects exert on their environment should not be included.

3. Choose a Set of Axes: Choose a set of x, y axes such that as many forces as possible lie directly along the x-axis or y-axis. It will minimize the number of forces to be resolved into components.

4. Resolve Forces into Components: Resolve all the forces which are not parallel to either of the axes into their rectangular components.

5. Apply Equilibrium Equations: Apply ΣFx=0and ΣFy=0 to get two equations.

6. Apply Torque Equation (if needed): If needed, apply Στ=0 to get another equation.

7. Solve the Equations: The equations can be solved simultaneously to find out the desired unknown quantities.

States of Equilibrium

An object is balanced when its center of mass and its point of support lie on the same vertical line. When forces on each side are balanced, the object is said to be in equilibrium. There are three states of equilibrium related to the stability of balanced bodies.

1. Stable Equilibrium:

A body is said to be in a state of stable equilibrium if, after a slight tilt, it comes back to its original position.

Explanation

Stable equilibrium occurs when the torques arising from the rotation (tilt) of the object compel the body back towards its equilibrium position.

Example

The cone shown in Figure 4.10 (a) is in a state of stable equilibrium. Its weight w, acting downward at the center of gravity G, and the reaction of the floor Fn, acting upward, lie on the same vertical line. Since these forces are equal and in opposite directions, they balance each other, satisfying both conditions of equilibrium.

When the cone is pushed slightly Figure 4.10(b), its center of gravity is raised but remains above the base. The weight w and the normal force Fn act like two unlike parallel forces, producing a clockwise torque that returns the cone to its original position.

Key Point The body remains in equilibrium as long as its center of mass lies within the base.

2. Unstable Equilibrium:

A body is in a state of unstable equilibrium if, after a slight tilt, it tends to moves further away from its original position.

Explanation

In unstable equilibrium, when the body is slightly disturbed, its center of mass no longer remains above the base, and the body topples over. The center of gravity lowers and continues to fall further, preventing the body from returning to its original position.

Example

Balancing a cone on its tip (Figure 4.11) is an example of unstable equilibrium. The weight w and the normal force Fn lie along the same line momentarily. However, even a slight tilt shifts the center of gravity outside the base, creating an anticlockwise torque that causes the cone to fall.

3. Neutral Equilibrium:

A body is in neutral equilibrium if it comes to rest in its new position after disturbances without any change in its center mass.

Explanation

In neutral equilibrium, tilting or moving the object does not create any torque to return it to its original position or move it further away. The center of mass remains at the same height.

Example

A cylinder resting on a horizontal surface (Figure 4.12) demonstrates neutral equilibrium. When the cylinder is rotated, the height of its center of mass remains unchanged, and the weight w and the ground's reaction force remain in the same vertical line. Other examples of neutral equilibrium are a ball rolling on a horizontal surface or a cone resting on its curved surface. (Figure 4.11).

Improvement of Stability

Stability is an essential concept in everyday life, and the position of the center of gravity plays a crucial role in determining it. Objects with a lower center of gravity are generally more stable, while those with a higher center of gravity are prone to instability.

Importance of Center of Gravity:

The position of the center of gravity significantly affects an object's stability. A low center of gravity ensures that the object remains in stable equilibrium. For example, a low armchair is more stable than a high chair because of its lower center of gravity. If disturbed slightly, the torque acting on the low chair brings it back to its original position.

Methods to Improve Stability

1. Lowering the Center of Gravity:

Placing heavier objects at a lower point within a system lowers the center of gravity, enhancing stability.

2. Widening the Base:

Increasing the base area provides better support and reduces the chances of tipping over, thereby improving stability.

Stability in Vehicles

The stability of vehicles, such as buses, depends on how they are loaded:

• Stable Loading:

When heavy loads are placed on the floor of the bus, its center of gravity remains low. In this condition, if the bus is disturbed slightly, a torque will bring it back to its original position. Therefore, the bus is in stable equilibrium.

• Unstable Loading:

If heavy loads, like steel sheets, are placed on the top of the bus, the center of gravity is raised. This makes the bus near a state of unstable equilibrium. A slight tilt could cause a couple that may turn it over.

Application in Ships and Boats:

Ships and boats follow similar stability principles. If heavy cargo is loaded at a lower level, the center of gravity remains low, improving stability. Conversely, placing heavy loads higher raises the center of gravity, making the vessel unstable and more likely to tip over.

Q.17. How is the concept of stability applied in real-life engineering, particularly in racing cars and balancing toys.

Ans Stability in Racing Cars

The concept of stability is widely applied to engineering technology, especially in manufacturing racing cars. As racing cars are driven at very high speeds and also have sharp turns in the track, the chances of the cars toppling over increase. To enhance the stability of racing cars, their centres of mass are kept as low as possible. Their base areas are also increased by keeping the wheels outside of their main bodies.

Stability in Balancing Toys

Balancing toys are also very interesting for both children and elders. The physics behind these toys is that stability is built into them. These toys are basically in a completely stable state, and their centres of gravity always remain below the pivot point. If the toys are disturbed in any direction, the centre of gravity is raised, and it becomes unstable for a moment. However, it comes back to its initial stable position by lowering its centre of gravity. The kids learn from these toys about stable systems and how they return to their

Stability in Racing Cars

The concept of stability is widely applied to engineering technology, especially in manufacturing racing cars. As racing cars are driven at very high speeds and also have sharp turns in the track, the chances of the cars toppling over increase. To enhance the stability of racing cars, their centres of mass are kept as low as possible. Their base areas are also increased by keeping the wheels outside of their main bodies.

Stability in Balancing Toys

Balancing toys are also very interesting for both children and elders. The physics behind these toys is that stability is built into them. These toys are basically in a completely stable state, and their centres of gravity always remain below the pivot point. If the toys are disturbed in any direction, the centre of gravity is raised, and it becomes unstable for a moment. However, it comes back to its initial stable position by lowering its centre of gravity. The kids learn from these toys about stable systems and how they return to their state of initial rest position after being disturbed. Educational games based on the principles of balancing toys have also been developed for kids, as shown in the figure.

Counterparts of Translational and Rotational Motion

Counterparts of velocity, acceleration, force, and momentum in translational motion are angular velocity, angular acceleration, moment of force (torque), and angular momentum respectively in rotational motion. It suggests that the torque plays the same role in rotational motion that is played by the force in translational motion.

Effect of Torque on Rotating Objects

Therefore, we are justified to predict that analogous to Newton's first law of motion, a rotating object will continue to do so with constant angular velocity unless acted upon by a resultant moment (torque). However, if a resultant torque is applied to a rotating object, it will accelerate depending on the direction of the torque relative to the axis of rotation.

This fundamental principle enhances our understanding of how objects move and interact with their environment whether in linear or rotational motion scenarios.

Velocity in Circular Motion

When a body is moving along a circular path, its velocity at any point is directed along the tangent drawn at that point. Figure 4.13 shows that the direction of the tangent at each point on a circle is different, therefore, the velocity of an object moving with uniform speed in a circle is changing constantly.

Role of Force

A force perpendicular to the direction of motion is always required to keep the object moving with uniform speed in a circular path. For instance, if it is not perpendicular to the velocity, the force (F) will have a component in the direction of velocity, which will change the magnitude of velocity. As the body moves with constant speed, this is possible only if the component of force along the velocity direction is zero, i.e., F cos 90° = 0.

Centripetal Force

We have studied above that an object can move in a circular path with uniform speed only if a force perpendicular to its velocity is acting constantly on it. This force is always directed towards the centre of the circle. It is called centripetal force and can be defined as:

"The force that causes an object to move in a circle at constant speed is called the centripetal force."

Formula for Centripetal Force

For an object of mass (m) moving with uniform speed v in a circle of radius r, the magnitude of centripetal force Fc , acting on it can be calculated by using the relation:

Fc = m v²/r

Sources of Centripetal Force (Real Life Examples):

We have learned that centripetal force has to be supplied if the body is to be maintained in its circular path. What could be the sources of centripetal force?

(i) Tension in a String: If we tie a stone to one end of a string and whirl it from the other end, we will have to exert a force on the stone through the string. If we release the string when it is at any point P, the stone will fly off along the tangent (PQ) to the circle. Then, it will move along the same straight line with constant velocity unless an unbalanced force acts upon it. In fact, the tension in the string was providing the stone the necessary centripetal force to keep it along the circular path. When we release the string, we stop applying force on the stone and hence it moves in a straight line.

(ii) Gravitational Force (Example of the Moon): Now consider the case of the moon, which moves around the Earth at constant speed. The gravity of the Earth provides the necessary centripetal force to keep it in its orbit. The same is the case of satellites orbiting the Earth in circular paths with uniform speed. The gravitational pull of the Earth provides the centripetal force.

(iii) Friction in a Washing Machine: One of the real-life examples is a washing machine dryer. A dryer is a metallic cylindrical drum with many small holes in its walls. Wet clothes are put in it. When the cylinder rotates rapidly, friction between clothes and drum walls provides the necessary centripetal force. As the water molecules are free to move, they cannot get the required centripetal force to move in circular paths and escape from the drum through the holes. This results in quick drying of clothes.

(iv) Centripetal Force in a Cream Separator: Another interesting example is that of a cream separator. In a cream separator, milk is whirled rapidly. The lighter particles of cream experience less centripetal force and gather in the central part of the machine. The heavier particles of milk need greater centripetal force to keep their circular motion in circles of small radius r. In this way, they move away towards the walls.